Let $f(x, y) = \tan(y^2) - 3xy$. Suppose $\vec{a} = (-1, 1)$ and $\vec{v} = \left( 1, 0 \right)$. Find the directional derivative of $f(x, y)$ at $\vec{a}$ in the direction of $\vec{v}$. $\dfrac{\partial f}{\partial v} = $
When a directional derivative is in the direction $(1, 0)$, $(0, 1)$, $(-1, 0)$, or $(0, -1)$, it becomes a regular partial derivative. Because $v = (1, 0)$, the directional derivative we want to find is also $\dfrac{\partial f}{\partial x}$ evaluated at $(-1, 1)$. $\begin{aligned} &\dfrac{\partial f}{\partial x} = -3y \\ \\ &\dfrac{\partial f}{\partial x}(-1, 1) = -3 \end{aligned}$ In conclusion, the directional derivative of $f$ at $\vec{a}$ in the direction of $\vec{v}$ equals $-3$.